# Topic :- Simple Interest (SI) and Compound Interest (CI)

By | September 18, 2015
Today’s topic is SIMPLE INTEREST & COMPOUND INTEREST

IMPORTANT FACTS AND FORMULAS

1. Principal: The money borrowed or lent out for a certain period is called the
principal or the sum.
2. Interest: Extra money paid for using other’s money is called interest.

3. Simple Interest (S.I.) : If the interest on a sum borrowed for a certain period is reckoned uniformly, then it is called simple interest.
Let Principal = P, Rate = R% per annum (p.a.) and Time = T years. Then,
(i) S.I. = (P*R*T )/100
(ii) P=(100*S.I)/(R*T) ;R=(100*S.I)/(P*T) and T=(100*S.I)/(P*R)

Compound Interest: Sometimes it so happens that the borrower and the lender agree to fix up a certain unit of time, say yearly or half-yearly or quarterly to settle the previous account.

In such cases, the amount after first unit of time becomes the principal for the second unit, the amount after second unit becomes the principal for the third unit and so on.

After a specified period, the difference between the amount and the money borrowed is called the Compound Interest (abbreviated as C.I.) for that period.

Let Principal = P, Rate = R% per annum, Time = n years.
I. When interest is compound Annually:
Amount = P(1+R/100)n
II. When interest is compounded Half-yearly:
Amount = P[1+(R/2)/100]2n¬
III. When interest is compounded Quarterly:
Amount = P[ 1+(R/4)/100]4n¬
IV. When interest is compounded AnnuaI1y but time is in fraction, say 3(2/5) years.
Amount = P(1+R/100)3 x (1+(2R/5)/100)
V. When Rates are different for different years, say Rl%, R2%, R3% for 1st, 2nd and 3rd year respectively.
Then, Amount = P(1+R1/100)(1+R2/100)(1+R3/100)
VI. Present worth of Rs.x due n years hence is given by :
Present Worth = x/(1+(R/100))n

SOLVED EXAMPLES
1. Find the simple interest on Rs. 68,000 at 16 2/3% per annum for 9 months.
P = Rs.68000,R = 50/3% p.a and T = 9/12 years = 3/4years.
S.I. = (P*R*T)/100 = Rs.(68,000*(50/3)*(3/4)*(1/100))= Rs.8500

2. Find the simple interest on Rs. 3000 at 6 1/4% per annum for the period from 4th Feb., 2005 to 18th April, 2005.
Time = (24+31+18)days = 73 days = 73/365 years = 1/5 years.
P = Rs.3000 and R = 6 ¼ %p.a = 25/4%p.a
S.I. = Rs.(3,000*(25/4)*(1/5)*(1/100))= Rs.37.50.
Remark : The day on which money is deposited is not counted while the day on which money is withdrawn is counted .

3. A sum at simple interests at 13 ½ % per annum amounts to Rs.2502.50 after 4 years find the sum.
Let sum be Rs. x then , S.I.=Rs.(x*(27/2) *4*(1/100) ) = Rs.27x/50
amount = (Rs. x+(27x/50)) = Rs.77x/50
77x/50 = 2502.50 x = 2502.50 * 50 = 1625
77
Hence , sum = Rs.1625.

4. A sum of Rs. 800 amounts to Rs. 920 in 8 years at simple interest rate is increased by 8%, it would amount to how much?
S.l. = Rs. (920 – 800) = Rs. 120; p = Rs. 800, T = 3 yrs.
. R = ((100 x 120)/(800*3) ) % = 5%.
New rate = (5 + 3)% = 8%.
New S.l. = Rs. (800*8*3)/100 = Rs. 192.
: New amount = Rs.(800+192) = Rs. 992.

5. Adam borrowed some money at the rate of 6% p.a. for the first two years , at the rate of 9% p.a. for the next three years , and at the rate of 14% p.a. for the period beyond five years. 1£ he pays a total interest of Rs. 11, 400 at the end of nine years how much money did he borrow ?
Let the sum borrowed be x. Then,
(x*2*6)/100 + (x*9*3)/100 + (x*14*4)/100 = 11400
(3x/25 + 27x/100 + 14x / 25) = 11400 95x/100 = 11400 x = (11400*100)/95 = 12000.
Hence , sum borrowed = Rs.12,000.

6. A certain sum of money amounts to Rs. 1008 in 2 years and to Rs.1164 in 3 ½ years. Find the sum and rate of interests.
S.I. for 1 ½ years = Rs.(1164-1008) = Rs.156.
S.l. for 2 years = Rs.(156*(2/3)*2)=Rs.208
Principal = Rs. (1008 – 208) = Rs. 800.
Now, P = 800, T = 2 and S.l. = 208.
Rate =(100* 208)/(800*2)% = 13%

7. At what rate percent per annum will a sum of money double in 16 years.
Let principal = P. Then, S.l. = P and T = 16 yrs.
Rate = (100 x P)/(P*16)% = 6 ¼ % p.a.

8. The simple interest on a sum of money is 4/9 of the principal .Find the rate percent and time, if both are numerically equal.
Let sum = Rs. x. Then, S.l. = Rs. 4x/9
Let rate = R% and time = R years.
Then, (x*R*R)/100=4x/9 or R2 =400/9 or R = 20/3 = 6 2/3.
Rate = 6 2/3 % and Time = 6 2/3 years = 6 years 8 months.

9. The simple interest on a certain sum of money for 2 l/2 years at 12% per annum is Rs. 40 less tban the simple interest on the same sum for 3 ½ years at 10% per annum. Find the sum.
Let the sum be Rs. x Then, ((x*10*7)/(100*2)) – ( (x*12*5)/(100*2)) = 40 (7x/20)-(3x/10)=40
x = (40 * 20) = 800.
Hence, the sum is Rs. 800.

10. A sum was put at simple interest at a certain rate for 3 years. Had it been put at 2% higher rate, it would have fetched Rs. 360 more. Find the sum.
Let sum = P and original rate = R.
Then, [ (P*(R+2)*3)/100] – [ (P*R*3)/100] = 360.
3PR + 6P – 3PR = 36000 6P=36000 P=6000
Hence, sum = Rs. 6000.

11. What annual installment will discharge a debt of Rs. 1092 due in 3 years at 12% simple interest?
Let each Instalment be Rs. x
Then, ( x+ ((x*12*1)/100)) + (x+ ((x*12*2)/100) ) + x = 1092
((28x/25) + (31x/25) + x) = 1092 (28x+31x+25x)=(1092*25)
x= (1092*25)/84 = Rs.325.
Each instalment = Rs. 325.

12. A sum of Rs. 1550 is lent out into two parts, one at 8% and another one at 6%. If the total annual income is Rs. 106, find the money lent at each rate.
Let the sum lent at 8% be Rs. x and that at 6% be Rs. (1550 – x).
((x*8*1)/100) + ((1550-x)*6*1)/100=106
8x + 9300 –6x=10600 2x = 1300 x = 650.
Money lent at 8% = Rs. 650. Money lent at 6%=Rs.(1550-650)= Rs. 900.

13. Find compound interest on Rs. 7500 at 4% per annum for 2 years, compounded annually.
Amount = Rs [7500*(1+(4/100)2] = Rs (7500 * (26/25) * (26/25)) = Rs. 8112. therefore, C.I. = Rs. (8112 – 7500) = Rs. 612.

14. Find compound interest on Rs. 8000 at 15% per annum for 2 years 4 months, compounded annually.
Time = 2 years 4 months = 2(4/12) years = 2(1/3) years.
Amount = Rs’. [8000 X (1+¬(15/100))2 X (1+((1/3)*15)/100)]
=Rs. [8000 * (23/20) * (23/20) * (21/20)]
= Rs. 11109. .
:. C.I. = Rs. (11109 – 8000) = Rs. 3109.

15. Find the compound interest on Rs. 10,000 in 2 years at 4% per annum, the
interest being compounded half-yearly.
Principal = Rs. 10000; Rate = 2% per half-year; Time = 2 years = 4 half-years.
Amount = Rs [10000 * (1+(2/100))4] = Rs(10000 * (51/50) * (51/50) * (51/50) * (51/50)) = Rs. 10824.32.
:. C.I. = Rs. (10824.32 – 10000) = Rs. 824.32.

16. Find the compound interest on Rs. 16,000 at 20% per annum for 9 months,
compounded quarterly.
Principal = Rs. 16000; Time = 9 months =3 quarters;
Rate = 20% per annum = 5% per quarter.
Amount = Rs. [16000 x (1+(5/100))3] = Rs. 18522.
CI = Rs. (18522 – 16000) = Rs. 2522.

17. If the simple interest on a sum of money at 5% per annum for 3 years is Rs. 1200, find the compound interest on the same sum for the same period at the same rate.
Clearly, Rate = 5% p.a., Time = 3 years, S.I.= Rs. 1200. . .
So principal=RS [100*1200]/3*5=RS 8000
Amount = Rs. 8000 x [1 +5/100]^3 – = Rs. 9261.
.. C.I. = Rs. (9261 – 8000) = Rs. 1261.

18. In what time will Rs. 1000 become Rs. 1331 at 10% per annum compounded annually?
Principal = Rs. 1000; Amount = Rs. 1331; Rate = 10% p.a. Let the time be n years. Then,
[ 1000 (1+ (10/100))n ] = 1331 or (11/10)n = (1331/1000) = (11/10)3
n = 3 years.

19. If Rs. 600 amounts to Rs. 683.20 in two years compounded annually, find the
rate of interest per annum.
Principal = Rs. 500; Amount = Rs. 583.20; Time = 2 years.
Let the rate be R% per annum.. ‘Then,
[ 500 (1+(R/100)2 ] = 583.20 or [ 1+ (R/100)]2 = 5832/5000 = 11664/10000
[ 1+ (R/100)]2 = (108/100)2 or 1 + (R/100) = 108/100 or R = 8
So, rate = 8% p.a.

20. If the compound interest on a certain sum at 16 (2/3)% to 3 years is Rs.1270,
find the simple interest on the same sum at the same rate and f or the same period.
Let the sum be Rs. x. Then,
C.I. = [ x * (1 + (( 50/(3*100))3 – x ] = ((343x / 216) – x) = 127x / 216
127x /216 = 1270 or x = (1270 * 216) / 127 = 2160.
Thus, the sum is Rs. 2160
S.I. = Rs ( 2160 * (50/3) * 3 * (1 /100 ) ) = Rs. 1080.

21. The difference between the compound interest and simple interest on a
certain sum at 10% per annum for 2 years is Rs. 631. Find the sum.
Let the sum be Rs. x. Then,
C.I. = x ( 1 + ( 10 /100 ))2 – x = 21x / 100 ,
S.I. = (( x * 10 * 2) / 100) = x / 5
(C.I) – (S.I) = ((21x / 100 ) – (x / 5 )) = x / 100
( x / 100 ) = 632 x = 63100.
Hence, the sum is Rs.63,100.

22. The difference between the compound interest and the simple interest accrued on an amount of Rs. 18,000 in 2 years was Rs. 405. What was the rate of interest p.c.p.a. ?
Let the rate be R% p.a. then,
[ 18000 ( 1 + ( R / 100 )2 ) – 18000 ] – ((18000 * R * 2) / 100 ) = 405
18000 [ ( 100 + (R / 100 )2 / 10000) – 1 – (2R / 100 ) ] = 405
18000[( (100 + R ) 2 – 10000 – 200R) / 10000 ] = 405
9R2 / 5 = 405 R2 =((405 * 5 ) / 9) = 225
R = 15.
Rate = 15%.

23. Divide Rs. 1301 between A and B, so that the amount of A after 7 years is equal to the amount of B after 9 years, the interest being compounded at 4% per annum.
Let the two parts be Rs. x and Rs. (1301 – x).
x(1+4/100)7 =(1301-x)(1+4/100)9
x/(1301-x)=(1+4/100)2=(26/25*26/25)
625x=676(1301-x)
1301x=676*1301
x=676.
So,the parts are rs.676 and rs.(1301-676)i.e rs.676 and rs.625.

24. A certain sum amounts to rs.7350 in 2 years and to rs.8575 in 3 years. Find the sum and rate percent.
S.I on rs.7350 for 1 year=rs.(8575-7350)=rs.1225.
Rate=(100*1225/7350*1)%=16 2/3%
Let the sum be rs.x.then,
X(1+50/3*100)2=7350
X*7/6*7/6=7350
X=(7350*36/49)=5400.
Sum=rs.5400.

25. A sum of money amounts to rs.6690 after 3 years and to rs.10,035 after 6 years on compound interest.find the sum.
Let the sum be rs.P.then
P(1+R/100)3=6690…(i) and P(1+R/100)6=10035…(ii)
On dividing,we get (1+R/100)3=10025/6690=3/2.
Substituting this value in (i),we get:
P*3/2=6690 or P=(6690*2/3)=4460
Hence,the sum is rs.4460.

26. A sum of money doubles itself at compound interest in 15 years.in how many years will it become eight times?
P(1+R/100)15=2P
(1+R/100)15=2P/P=2
LET P(1+R/100)n=8P
(1+R/100)n=8=23=(1+R/100)153[USING (I)]
(1+R/100)N=(1+R/100)45
n=45.
Thus,the required time=45 years.

27.What annual payment will discharge a debt of Rs.7620 due in 3years at
16 2/3% per annum interest?
Let each installment beRs.x.
Then,(P.W. of Rs.x due 1 year hence)+(P>W of Rs.x due 2 years hence)+(P.W of Rs. X due 3
years hence)=7620.
x/(1+(50/3*100))+ x/(1+(50/3*100))2 + x/(1+(50/3*100))3=7620
(6x/7)+(936x/49)+(216x/343)=7620.
294x+252x+216x=7620*343.
x=(7620*343/762)=3430.
Amount of each installment=Rs.3430.